This question is a hypothesis testing problem for the population mean. Here's how we can approach it:

Given:

• Population mean ($\mu_0$) = 82
• Population standard deviation ($\sigma$) = 20
• Sample size ($n$) = 81
• Sample mean ($\bar{x}$) = 90
• Significance level ($\alpha$) = 0.05

We will use the z-test for the hypothesis test because the population standard deviation is known, and the sample size is large (greater than 30).

Step 1: Formulate Hypotheses

• Null hypothesis ($H_0$): $\mu = 82$ (The mean score is 82)
• Alternative hypothesis ($H_1$): $\mu \neq 82$ (The mean score is not 82)

This is a two-tailed test.

Step 2: Compute the z-test statistic

The z-test statistic is given by the formula:

$z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}$

Where:

• $\bar{x}$ = 90 (sample mean)
• $\mu_0$ = 82 (population mean)
• $\sigma$ = 20 (population standard deviation)
• $n = 81$ (sample size)

Step 3: Critical z-value

For a significance level of $\alpha = 0.05$ and a two-tailed test, the critical z-values are $\pm 1.96$.

Let me calculate the z-value next.The computed z-test statistic is approximately 3.60.

Step 4: Conclusion

Since the computed z-value of 3.60 exceeds the critical z-value of ±1.96 (for a significance level of 0.05), we reject the null hypothesis.

Interpretation:

There is enough evidence at the 0.05 confidence level to reject the claim that the mean score of the students is 82. The sample mean of 90 is significantly different from 82.

Would you like further details or have any questions?

Related questions:

1. What is a two-tailed hypothesis test?
2. How is the significance level (α) determined in hypothesis testing?
3. What are the conditions for using a z-test versus a t-test?
4. How would the result change if the sample size were smaller?
5. How does changing the confidence level affect hypothesis test results?

Tip:

Always verify the assumptions of normality and known standard deviation before applying a z-test in hypothesis testing.